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18.099 Final Project # 5
The Cartan matrix of a Root System
Thomas R. Covert July 30, 2004
A root system in a Euclidean space V with a symmetric positive definite inner product �,� is a finite set of elements Δ of V suchthat
1.
Δ spans V;
2.
for every root α∈ Δ and every β∈ Δ, β− 2�β,α� αis a root. Moreover,
�α,α�
every root has such an expansion;
3. the number 2�β,α� is an integer for all α,β∈ Δ.
�α,α�
If 2α�∈ Δ for all α∈ Δ, the rootsystem Δ is reduced.For everyroot system in V, there exists a simple root system Π ⊂ V, such that
1.
the elements of Π form a basis for V;
2.
every root β∈ Δ is a linear combination πi∈Π ciπi with every ci being of the same sign.
If the coefficients ci for a root β are all nonnegative, the root is positive. Otherwise, it is negative. The set of positive roots in Δ is denoted as Δ+ .
Definition 1 Let Π ∈ Δ be a simple root system in V, and let the elements of Π be enumerated as {αi}n where n is the dimension of V. The Cartan
i=1
matrix Ais the square matrix given by Aij =2�αi,αj�
.
�αi,αi�
1
� �
It is obvious that this matrix is dependent on the enumeration of the elements of Π.
However, the Cartan matrices of a root system with different enumerations of the simple roots are related by permutation matrices. To prove this, an important fact about permutation matrices is necessary
Theorem 2 Let Pij be the identity matrix with rows i and jswitched. For any square matrix B, the matrix B� = PijB(Pij)−1 is B with rows and columns iand jswitched.
Proof: Because Pij is the identity matrix with rows i and j switched, the lefthand product PijB is the matrix Br,
which is B with rows i and j switched. The transpose (Br)� = B�(Pij)� is the transpose of B with columns i and j switched. Since Pij is a simple permutation matrix (i.e., it is only one row exchange away from the identity matrix), (Pij)−1 =
(Pij)�. B�(Pij)−1
Therefore, (Br)� = . Multiplying (Br)� on the left by
the same Pij will again switch rows iand j. This results in the matrix B�� = Pij(Br)�
= PijB�(Pij)−1, which is the transpose of B with the ith and PijB(
Pij)−1jth rows and columns switched.
Its transpose, B� =(B��)� =,
is therefore B with the ith and jth rows and columns switched.
2 With this theorem, it is now possible to relate the Cartan matrices given by different enumerations of the same reduced root system.
Corollary 3 Let A be the Cartan matrix of a reduced root system (Π,Δ) with a fixed enumeration of the simple roots {αi}n and let A� be the Cartan
i=1
matrix of the same root system with the same enumeration of the simple
PklA(Pkl)−1
roots, except that roots αk and αl are reversed. Then A� = .
Proof: The Cartan matrix for the first enumeration is given by
Aij =2�αi,αj�
.
�αi,αi�
For all entries (i,j), i,j= k and i,j= l, the Cartan matrix A�
ij = Aij. All
entries Akj are given by 2�αl,αj� , and the entries Alj,Aik,Ail are given in a
�αl,αl�similar manner. Thus, the entries of A� involving k are switched with those involving l and vice versa. Then A� is A with rows and columns i and j PijA(Pij)−1
switched. By Theorem 2, A� =. 2 2
�
Because any enumeration of a set of simple roots is related to any other enumeration by some finite number of permutations, the relationship between any two Cartan matrices for a root system is an isomorphism by the conjugate product of permutation mtrices.
Definition 4 Two Cartan matrices are isomorphic if they are conjugate by a product of permutation matrices.
Here are some examples of Cartan matrices.
Example 5 The Cartan matrix of the root system of type An as defined in Final Project 3 is the n×n tridiagonal matrix with 2’s on the main diagonal and −1’s on the upper and lower diagonals.
Proof: Let the root system of type An be as defined in [3]. The simple
n+1
roots Π are enumerated as {ei −ei+1}i=1 so that there are n distinct simple roots. For any two roots ei −ei+1 and ei+1 −ei+2, the appropriate entry of Ais
2�ei −ei+1,ei+1 −ei+2�
.
ei −ei+1,ei −ei+1�
The numerator of that fraction is
ei+1,ei+1�+ �ei+1,ei+2�
�ei,ei+1�−�ei,ei+2�−�and since all of the e’s are orthonormal to each other and of length 1, the numerator reduces to −1. The denominator is
ei −ei+1,ei −ei+1�= ei+1,ei�+ �ei+1,ei+1�
��ei,ei�−�ei,ei+1�−�
which reduces to 2 for the same reasons. Hence, the value of an entry directly above or below the diagonal in this Cartan matrix is always −1. The diagonal entries are always 2,
because the inner products in the numerator and denominator are identical. All other entries are 0 because the inner product in the numerator involes no identical terms. Therefore, the Cartan matrix A of the root system of type An is given by
0 i−j>1,
||
Aij = −1 |i−j=1,
|
2 i−j=0.
||
⎧ ⎪⎨ ⎪⎩
3
2
�
�
An interesting fact about this Cartan matrix is that it is identical to the stiffness matrix of a system with n−1 springs with unit spring constants and n unit masses.
Example 6 Let the root system of the type B4 be defined as in [1]. The simple roots Π are {e1 −e2,e2 −e3,e3 −e4,e4}. The positive roots Δ+ are
4
ei ±ej}i{
⎞
⎛
⎜⎜⎝
2 −10 0 2 −10
−1 02
⎟⎟⎠
.
0
−1
0
−2
−1
2
�4 Proof: These roots form a basis for R4 . For any vector v ∈ R4 = ciei, let c1 = b1,c2 = b2 −b1,c3 = b3 −b2, and c4 = b4 −b3. The
i=1
sum then expands to b1(e1 − e2)+ b2(e2 − e3)+ b3(e3 − e4)+ b4e4. It is clear, then, that the simple roots in B4 are a basis for R4 and thus they span it. Additionally, every δ ∈ B4 has an integral expansion in the sim4
ple roots. Recall that B4 = {ei ±ej}iei}i=1. Let each simple root
=j ∪{±
enumerated above be denoted by πi. The roots ei can be expressed as
�4 ±
the sum �4 j=1 πj. Therefore, the roots ±ei �ej can be expressed as the
± �4
sum ± k=1 πk + � l=1 πl. In this way, all roots are combinations of the simple roots with integer coefficients of the same sign. The positive roots are those with all positive coefficients in their expansion across the simple roots. Clearly every ei is a positive root, and thus every sum ei + ej,j= iis
also a positive root. Because every simple root but the last one is identical to the simple roots of A4, the Cartan matrix is identical except in the last subdiagonal entry,
where it is −2. This is because the square of the length of e4 is half of the square of the length of e3 −e4.
The properties of all Cartan matrices are summarized here:
Theorem 7 Any Cartan matrix Aof a root system (Π,Δ) has the following properties.
1.
every entry is an integer;
2.
all diagonal entries are 2;
3.
all off diagonal entries are nonpositive;
4
2
� �
4.
Aij =0 if and only if Aji =0;
5.
there exists a diagonal matrix D with positive entries such that DAD−1 is symmetric positive definite.
Proof: 1. This is true by property 3 of the definition of an abstract root system [1].
2. Every entry Aii = 2�αi,αi� = 2.
�αi,αi�
3. Let αi and αj be two distinct simple roots. By property 3 of root systems, αi − 2�αi,αj� αj is a root αk. Since αi and αj are simple, and αk
�αj,αj
is a linear combination of the two, property 2 of simple roots requires that the number 2�αi,αj� is negative.
�αj,αj
4. Suppose Aij = 0. Then �αi, αj� = 0. By the reflexivity of the inner product, �αi, αj� = �αj, αi�. Therefore, if the numerator of Aij =0, the numerator of Aji must also be 0 and vice versa.
|
=
�αi, αi�
1
2
.
Then DAD−1
ij
5. Let D = diag(α1,α2, ..., αn|)where αi
|
|
|
|
|
|
=
1
2
1
2
Aij �αi,αi�
�αj,αj�
2�αi,αj�
1
2�αi,αj�
this product reduces to DAD−1
ij
Because Aij = =
.
�αi,αi�
which, by the reflexivity of the inner product, is clearly
1
�αi,αi�
symmetric.
2
�αj,αj�
2
Denote the jth entry of the simple root αi in some basis The matrix B defined =
bij
of V as bij.
by Bij
gives a Cholesky
1
2
�αi,αi�
factorization for A because B�Bij is the inner product of αi with the αj
divided by the lengths of αi and αj. This is precisely DAD−1 . Because
ij
the simple roots form a basis for V , B is invertible. Thus A is positive definite.
Example 8 There are 5 reduced root systems in R2: A1 ⊕ A1, A2, B2, C2, and G2:
1. The root system A1 ⊕ A1 consists of {±e1, ±e2} so that the angle between the simple roots {
e1, e2} is π and each simple root has unit length.
2
The off diagonal entries of the Cartan matrix must be 0 because the
2 0
simple roots are orthogonal. Therefore, the Cartan matrix is
0 2
5
2
� �
� � � �
� �
�
� �
and since the Cartan matrix is already symmetric and positive definitie, the diagonalizing matrix is the identity matrix.
2. The root system A2 consists of 6 vectors arranged in a hexagonlike
fashion and the angle between the two simple roots is
π, each of unit 3
length. The Cartan matrix of this root system is thus 2 −1 and
2
−1
since the Cartan matrix is already symmetric and positive definite, the diagonalizing matrix is the identity matrix.
3. The simple roots of the root system B2 are {e1 − e2, e2}, resulting in 3and root lengths of 1 and √2. Thus the Cartan matrix
π 4
� is
2 −2
� −1 2
. The diagonalizing matrix is therefore � �
��√2 0 0 1
. The
2
an angle of
−√2
product DAD−1 = −√2 2 has determinant 2 so DAD−1 is symmetric positive definite.
π 4
4. The simple roots of the root system C2 are {e2 − e1,2e1}. Thus, the 3and their lengths are √2 and 2. The Carangle between them is
tan matrix is
2 −1
−2 2
and the diagonalizing matrix D is � �
2 0 0 √2
2
−√2
so that the product DAD−1 = −√2 2 is symmetric positive definite. Note that the diagonalizing matrix for C2 is just √2 times the diagonalizing matrix for B2. In that sense, these root systems are equivalent.
5. The simple roots of the root system G2 have lengths √3 and 1 and the
π 6
angle between them is 5
�√3 0
the diagonalizing matrix is
2 −1
. Therefore, the Cartan matrix is
−3
,
, and the symmetric positive def01
2 −√3
−√32
inite product of the two is
.
Theorem 9 Areducedroot system is reducibile if and only if for some choice of a simple root system and some enumeration of the ndices of the simple roots, the Cartan matrix is block diagonal with more than one block.
6
2
Proof: Suppose that a root system is reducible as Δ =Δ� ∪Δ�� so that any
n
δ� ∈Δ� is orthogonal to any δ�� ∈Δ��. Let the simple roots of Δ be {αi}i=1
rn
so that {αi}i=1 ∈Δ� and {αi}i=r+1 ∈Δ��. Then for any i ∈{1, 2, .., r}and r + 1, r + 2, ..., n} (i.e., i and j represent roots from different parts
j ∈ {
of Δ’s decomposition), Aij = 0. Then clearly A is block diagonal with 2 blocks. If a root system is reducible into n orthogonal components, by the same argument, it will have n blocks.
Now, assume that the Cartan matrix of a root system Δ is block diagonal
r
with 2 blocks. Let the roots corresponding to the first block be {αi}i=1 and
n
those in the second block be {αi}i=r+1. Assume α is a positive root in Δ. By theorem 7 of [2], α = a1 + a2 + ... + ak, where each ai is a simple root, and repetitions are allowed. Furthermore, each partial sum a1 + a2 + ... + ac is a root, for all c ≤k. Let B be the cth partial sum for α and let A be the next simple root in the sum for α.
Then B and B + A are both roots. Theorem 10 of [1] requires that for all n such that B + nA is a root or 0, there exists positive p and q suchthat −p ≤n ≤q and p −q = 2�B,A� .
�B,B�
Suppose the roots that sum to B and A are in different blocks, meaning that �B, A�= 0 and p = q. Then if B + A is a root (n =1 in the theorem from [1]), B −A must also be a root. However, this results in a root which is a sum of simple roots with mixed sign coefficients. This contradicts the definition of simple roots. Therefore, the only such p that does not result in this contradiction is p = 0. This means that in any expansion of a positive root in a root system whose Cartan matrix has 2 blocks, all of the coefficients for the set of simple roots in one orthogonal component must be 0.
2
If there is only one block, the root system is said to be irreducible.
Theorem 10 The matrix D which diagonalizes the Cartan matrix of a root system is determined uniquely up to a scalar multiple oneach block of A.
Proof: For every pair αi and αj that are in different blocks, the entry Aij of the Cartan matrix is 0. Therefore, any diagonal matrix D symmetrizes such entries in the Cartan matrix A.For any matrix D� = diag(a1, a2, ..., an) that symmetrizes A, and any two simple roots αi and αj which are in the
ai aj
same block, D�A(D�)−1 = Aijaj . Because D� symmetrizes A, Aij ai = Ajiai .
ij aj aj αi2
2
This reduces to ai = |αj|2and finally to a2 = |αj|2 aj. This implies that
aj ai i|αi|2||
ai and aj are determined uniquely up to a scalar multiple by the relative lengths of αi and αj. 2 7
With this theorem, it is clear that for any irreducible root system, the matrix D gives the relative lengths of the roots. By definition, the entries of the Cartan matrix Agive the angles between any two simple roots. Then, Aand Dcompletely characterize the simple roots.
Corollary 11 Any Cartan matrix determines a set of simple roots uniquely up to a scalar multiple of an orthogonal transformatio on each irreducible component
Proof: From Theorem 10 it is clear that given a Cartan matrix A, there is a diagonalizing matrix D which is unique up to a scalar multiple in each block of A. Thus, A itself determines the relative lengths of some set of simple roots (up to a scalar multiple on each block). Additionally, every entry (i,j) of A gives the angle between αi and αj. So for each irreducible component of A, there is a unique root system.
Example 12 The root system A1 ⊕ B2 has {e1,e2 − e3,e3} as simple roots. It’s Cartan matrix is ⎛⎞
20 0 ⎝02 −1⎠ 2
0 −2
which is block diagonal with the Cartan matrix for A1 in the first block, and the Cartan matrix for B2 in the second.
References
[1] Dr. Anna Lachowska, Final Project 2: Abstract Root Systems, preprint, MIT, 2004.
[2] Dr. Anna Lachowska, Final Project 4: Properties of Simple Roots, preprint, MIT, 2004.
[3] Dr. Anna Lachowska, Final Project 3: Properties of Simple Roots, preprint, MIT, 2004.
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